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MLB (Nationals @ Marlins): Will Stephen Strasburg (WSH) record a STRIKEOUT in EACH of the FIRST TWO INNINGS?


7:00PM ESPN
Yes: Strasburg K in each of first two innings
No: Strasburg no K in at least 1 of those innings


Inputs To Solve

Strike Outs per Innings by Strasburg

##### User Estimates #####
KperInning = 120/102

print("Use expected Ks per Inning as %s." % (round(KperInning,2)))

Use expected Ks per Inning as 1.18.     


Method to Solve

lambda_ = KperInning

print("lambda = the total expected Ks recorded by Strasburg in one inning")
print("lambda ~ %s" % (round(lambda_,2)))

lambda = the total expected Ks recorded by Strasburg in one inning
lambda ~ 1.18     

import math
str_ = ""
print("The probability of k events occurring in a Poisson interval = e^(-lambda) * (lambda^k)/k!")
print(' ')

p = 0
for i in range(0,4):
    p += math.exp(-lambda_)*(lambda_**(i))/(math.factorial(i))
    
    print("e^(-%s) * (%s^%s)/%s!" % (round(lambda_,2),round(lambda_,2),(i),(i)))
    if i == 0:
        p_0 = math.exp(-lambda_)*(lambda_**(i))/(math.factorial(i))
        print(round(p_0,3))
        print('')

print(' ')
print(str_)
print("The probability that there are 3 or less Ks in an inning is %s, (this is the scaling factor)" % round(p,3))

The probability of k events occurring in a Poisson interval = e^(-lambda) * (lambda^k)/k!

e^(-1.18) * (1.18^0)/0!
0.308

e^(-1.18) * (1.18^1)/1!
e^(-1.18) * (1.18^2)/2!
e^(-1.18) * (1.18^3)/3!


The probability that there are 3 or less Ks in an inning is 0.968, (this is the scaling factor)     

scaled_p = p_0/p

print("The probability Strasburg records zero Ks in any inning is %s" % round(scaled_p,3))

The probability Strasburg records zero Ks in any inning is 0.318     


Solution

p_both = (1-scaled_p)**2
print("The probability that Stephen Strasburg records a STRIKEOUT in EACH of the FIRST TWO INNINGS is ~%s" % (round(p_both,3)))

The probability that Stephen Strasburg records a STRIKEOUT in EACH of the FIRST TWO INNINGS is ~0.464

Posted on 6/27/2019






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